4
$\begingroup$

The Birman-Schwinger principle says that if $\Delta$ is the usual Laplacian on $\mathbb{R}^n$ and we consider the operator $H=-\Delta-V$ for a positive potential $V$, then, for any $\lambda>0$, the number of eigenvalues at most $-\lambda$ of this operator is the same as the number of eigenvalues at least 1 of the operator

$$ K_\lambda=V^{1/2}(-\Delta+\lambda)^{-1}V^{1/2}.$$

Can you suggest any source where this result is proved for $\lambda=0$?

Many thanks!

Improve this question
$\endgroup$
2
  • $\begingroup$ What is $(-\Delta)^{-1}$? $\endgroup$
    – Fedor Petrov
    Commented Dec 13, 2016 at 21:15
  • $\begingroup$ We can define it for example via $\mathcal{F}((-\Delta)^{-1}f)(x)=|x|^{-2}\mathcal{F}(f)(x)$, where $\mathcal{F}$ is just Fourier transform. $\endgroup$
    – cav
    Commented Dec 13, 2016 at 23:33

1 Answer 1

Reset to default
5
$\begingroup$

This runs into obvious technical issues. For example, $K_0$ will not be bounded (let alone compact) even for very nice $V$. So one also has to think about what exactly one wants to prove.

This paper discusses these issues. In particular, the Birman-Schwinger principle for $\lambda=0$ is stated in equation (1.6).

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .